--- title: "HR trickling filter, 2.5 MLD; change if standard" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:trickling-hr-2p5mld-std-compare" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-hr-2p5mld-std-compare site: BCT Engineering --- # HR trickling filter, 2.5 MLD; change if standard Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2076 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-hr-2p5mld-std-compare - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Determine the dimensions of a high rate trickling filter for the following data: (i) Sewage flow = 2.5 MLD; (ii) Recirculation ratio = 1.5; (iii) BOD of raw sewage = 300 mg/l; (iv) BOD removal in primary settling tank = 30%; (v) Final effluent BOD desired = 35 mg/l. By what % the diameter of the filter will have to be modified if it is to be designed as a standard rate trickling filter for above requirements? ## Short answer High-rate filter volume 1363.8 m³, depth 2.0 m, diameter 29.5 m. If designed as standard-rate, diameter must be increased by 37.5% (to 40.5 m). ## Answer #### Data and approach High-rate filter design uses the **NRC (National Research Council) equation**; $W$ is in kg/day of BOD applied to the filter, $V$ in m³: $$ E = \frac{100}{1+0.4432\sqrt{\dfrac{W}{VF}}},\qquad F=\frac{1+R}{(1+0.1R)^2} $$ #### Step 1: BOD applied to the filter and required efficiency $$ \begin{aligned} Q &= 2500\ \text{m}^3/\text{day}\\ \text{BOD after primary} &= 300\times(1-0.3) = 210.0\ \text{mg/l}\\ W &= Q\times BOD = 2500\times 210.0\times 10^{-3} = 525.0\ \text{kg/day}\\ E &= \frac{210.0-35}{210.0}\times 100 = 83.33\% \end{aligned} $$ #### Step 2: Recirculation factor $$ F = \frac{1+1.5}{(1+0.1\times 1.5)^2} = 1.890 $$ #### Step 3: Volume $$ 83.33 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow s=\sqrt{\frac{W}{VF}} = 0.4513 \Rightarrow VF = \frac{525.0}{0.4513^2} = 2578.1 $$ $$ V = \frac{VF}{F} = \frac{2578.1}{1.890} = 1363.8\ \text{m}^3 $$ #### Step 4: Diameter Adopt depth $d = 2.0$ m (high-rate filters are 1.8-3 m deep). $$ A = \frac{V}{d} = 681.9\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 29.47\ \text{m (adopt 29.5 m)} $$ #### Step 5: Checks $$ \begin{aligned} \text{Organic loading} &= \frac{W}{V} = 0.38\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.32-1.0, OK})\\ \text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 9.2\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40; just below, acceptable}) \end{aligned} $$ **Answer: single-stage high-rate filter of volume about 1363.8 m³, depth 2.0 m, diameter 29.5 m.** #### Change in diameter if designed as a standard-rate filter A standard-rate filter has no recirculation ($R=0$, $F=1$). The same efficiency $E=83.33\%$ (same $s=0.4513$) then needs $$ V_{std}=\frac{W}{s^2}=\frac{525.0}{0.2036}=2578.1\ \text{m}^3 $$ With the same depth of 2.0 m, $D_{std}=\sqrt{4V_{std}/(\pi\times2)}=40.5$ m, compared with $D=29.5$ m for the high-rate filter. $$ \text{Change in diameter}=\frac{40.5-29.5}{29.5}\times100=37.5\%\ (\text{since } D\propto\sqrt{V}\propto\sqrt{F}=\sqrt{1.890}) $$ **The diameter must be increased by about 37.5%** if the filter is built as a standard-rate unit. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With raw BOD 300 mg/l, 30% PST removal and final BOD 35 mg/l, what filter efficiency is required? A. 88.3% B. 70% C. 83.33% D. 66.7% Answer: C. Settled BOD = 210 mg/l; E = (210-35)/210 = 83.33%. (Key point: Efficiency required) 2. For R = 1.5 what is F, and what does it do to the required volume? A. F = 1.890; the volume from the NRC equation is VF/F, so recirculation reduces the volume B. F = 2.5; volume increases by 2.5 times C. F = 1.890; volume = VF x F D. F = 1.5; volume is unaffected Answer: A. Recirculation raises F, so for the same VF the required V = VF/F is smaller. (Key point: Recirculation factor) 3. How is the standard-rate filter treated for the same efficiency? A. Keep R = 1.5 but halve the depth B. Take R = 0 so F = 1; the volume then equals VF = 2578.1 m3 C. Use R = 1 so F = 1 D. Use the same volume 1363.8 m3 with a deeper tank Answer: B. A standard-rate filter has no recirculation, so F = 1 and V_std = W/s^2 = 2578.1 m3. (Key point: Standard-rate equivalent) 4. Since D is proportional to sqrt(V) at equal depth, by what percentage must the diameter change for a standard-rate filter? A. Decrease by 37.5% B. Increase by 89% C. Increase by 11% D. Increase by about 37.5% (sqrt(1.890) - 1) Answer: D. V_std/V = F = 1.890, so D_std/D = sqrt(1.890) = 1.375, i.e. 37.5% larger (40.5 m vs 29.5 m). 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